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Two crickets, Chirpy and Milada, jump from the top of a vertical cliff. Chirpy just drops and reaches the ground in 3.50 s, while Milada jumps horizontally with an initial speed of 95.0 \(\mathrm{cm} / \mathrm{s} .\) How far from the base of the cliff will Milada hit the ground?

Short Answer

Expert verified
Milada will hit the ground 3.325 meters from the cliff.

Step by step solution

01

Convert Units and Gather Known Values

First, convert Milada's horizontal speed from centimeters per second to meters per second. Given speed is 95.0 cm/s, which is equivalent to 0.95 m/s. Next, note the time it takes for both crickets to reach the ground, which is 3.50 seconds.
02

Calculate Milada's Horizontal Distance

Since Milada's horizontal velocity is constant and there is no horizontal acceleration, use the formula to find the horizontal distance: \( d = v \cdot t \). Here, \( v = 0.95 \text{ m/s} \) and \( t = 3.50 \text{ s} \). Thus, \( d = 0.95 \times 3.50 = 3.325 \text{ m} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Horizontal Motion
When analyzing projectile motion, it's important to split it into two main components: horizontal and vertical motion. Horizontal motion refers to any motion that occurs along the horizontal axis. In our example, we look at Milada's leap. Milada jumps with a horizontal velocity of 0.95 m/s. This movement is at a constant speed as there are no external horizontal forces acting on it once the jump is made. This means no horizontal acceleration is involved.
Milada's horizontal movement is governed by a simple formula, which is the product of velocity and time. By using the formula \( d = v \cdot t \), where \( d \) represents distance, \( v \) is velocity, and \( t \) is time, we can assess how far Milada moves horizontally. Understanding this principle helps solve problems where objects are projected horizontally.
Vertical Drop
Let's delve into the vertical aspect of the motion. The vertical drop is influenced by gravity, which works to pull the cricket straight down towards the ground. In this case, both Chirpy and Milada experience the same vertical drop, landing after 3.50 seconds. Gravity causes them to accelerate downwards at around 9.81 m/s², assuming negligible air resistance.
In projectile problems like this, we typically separate the vertical and horizontal calculations. The vertical element doesn't influence the horizontal movement, and this principle underpins the independent motion of the vertical path. Thus, Chirpy and Milada drop to the ground in the same amount of time, showcasing the key concept that horizontal velocity does not affect the time taken to fall a certain vertical distance.
Conversion of Units
In physics, correct unit conversion is critical to solving problems accurately. The exercise provided involves converting Milada's initial horizontal velocity from centimeters per second to meters per second. This conversion is necessary as most physics equations utilize standard units — meters in this case.
To convert 95.0 cm/s to m/s, divide by 100 (since there are 100 centimeters in a meter). Thus, 95.0 cm/s becomes 0.95 m/s.
  • This is an essential step before calculations as mixing units can lead to errors.
  • Always double-check your conversions to maintain consistency and accuracy in your results.
  • Conversion is not just a mathematical procedure, but a fundamental part of understanding and communicating scientific data effectively.
Time of Flight
The time of flight refers to the duration that an object remains in the air before it lands. In the example of the crickets, both Chirpy and Milada are in flight for 3.50 seconds.
This time is determined by the vertical component of motion — namely gravity and the height from which they fall. Since both crickets start from the same height, and gravity acts uniformly on both, they share this flight time.
Understanding the time of flight helps predict and understand both where and when an object, such as Milada, will land once it is airborne.
  • Time of flight is pivotal in determining other aspects, like the horizontal distance covered.
  • Knowledge of this time is useful in solving problems related to motion, as it reveals insights into the influence of gravity and initial launch conditions.
Knowing how to calculate and apply time of flight enhances one's ability to analyze and interpret projectile motion effectively.

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Most popular questions from this chapter

A projectile is being launched from ground level with no air resistance. You want to avoid having it enter a temperature inversion layer in the atmosphere a height \(h\) above the ground. (a) What is the maximum launch speed you could give this projectile if you shot it straight up? Express your answer in terms of \(h\) and \(g .\) (b) Suppose the launcher available shoots projectiles at twice the maximum launch speed you found in part (a). At what maximum angle above the horizontal should you launch the projectile? (c) How far (in terms of \(h\) ) from the launcher does the projectile in part (b) land?

A shot putter releases the shot some distance above the level ground with a velocity of \(12.0 \mathrm{m} / \mathrm{s}, 51.0^{\circ}\) above the horizontal. The shot hits the ground 2.08 s later. You can ignore air resistance. (a) What are the components of the shot's acceleration while in flight? (b) What are the components of the shot's velocity at the beginning and at the end of its trajectory? (c) How far did she throw the shot horizontally? (d) Why does the expression for \(R\) in Example 3.8 not give the correct answer for part \((\mathrm{c}) ?\) (e) How high was the shot above the ground when she released it? \((\mathrm{f})\) Draw \(x-t\) , \(y-t, v_{x}-t,\) and \(v_{y}-t\) graphs for the motion.

A small toy airplane is flying in the \(x y\) -plane parallel to the ground. In the time interval \(t=0\) to \(t=1.00 \mathrm{s}\) its velocity as a function of time is given by \(\vec{v}=\left(1.20 \mathrm{m} / \mathrm{s}^{2}\right) t \hat{\imath}+\left[12.0 \mathrm{m} / \mathrm{s}-\left(2.00 \mathrm{m} / \mathrm{s}^{2}\right) t\right] \hat{\boldsymbol{J}}\) . At \(\quad\) what value of \(t\) is the velocity of the plane perpendicular to itsM acceleration?

A 124 -kg balloon carrying a 22 -kg basket is descending with a constant downward velocity of 20.0 \(\mathrm{m} / \mathrm{s} . \mathrm{A} 1.0\) -kg stone is thrown from the basket with an initial velocity of 15.0 \(\mathrm{m} / \mathrm{s}\) perpendicular to the path of the descending balloon, as measured relative to a person at rest in the basket. The person in the basket sees the stone hit the ground 6.00 s after being thrown. Assume that the balloon continues its downward descent with the same constant speed of 20.0 \(\mathrm{m} / \mathrm{s} .\) (a) How high was the balloon when the rock was thrown out? (b) How high is the balloon when the rock hits the ground? (c) At the instant the rock hits the ground, how far is it from the basket? (d) Just before the rock hits the ground, find its horizontal and vertical velocity components as measured by an observer (i) at rest in the basket and (ii) at rest on the ground.

An airplane is flying with a velocity of 90.0 \(\mathrm{m} / \mathrm{s}\) at an angle of \(23.0^{\circ}\) above the horizontal. When the plane is 114 \(\mathrm{m}\) directly above a dog that is standing on level ground, a suitcase drops out of the luggage compartment. How far from the dog will the suitcase land? You can ignore air resistance.

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